1.5v to 5v

[pyrohaz1]

Howdy guys, hope you all had a good Christmas and new year, I haven't been here in ages! Right, i've come across the question of... can somebody help me in designing a circuit that can complete my specification:

DISCREET! No maxim or national semiconductor IC's please.
1.5v to 5v, no more than 1A input current
5v output current at 0.2A (preferably 0.5A but not urgent)
Assuming 1A input current and Output current of 0.2A, hopefully an efficiency above 66%?
Output can be zener regulated or shunt regulated etc...
Hopefully using boost converter (inductor based) circuitry OR if anybody can devise a high current switched capacitor circuit, any ideas are welcome! I originally designed one using a transistor multivibrator but when testing in real life, it sagged to 3v with a 100R load!

I was thinking, a pulse width modulating RTL inverter style thing, I will attach a schematic for the sort of oscillator i

View attachment 40831

 

[audioguru2]

I don't think your oscillator will have a symmetrical output so it will not be a square-wave.
Use a Cmos IC oscillator instead.
A CD4047 Cmos IC has a RC oscillator followed by a digital divider that produces a perfect square-wave plus a second output that is the opposite (an inverted second output).

 

[pyrohaz1]

Since the 4047 is in the 4000 series, I assume it doesn't work down to 1.5v? I thought their voltage range was just 3v-15v? Also, I thought that with boost converters, if the duty cycle is higher than 50%, the efficiency can be increased by matching the duty cycle to the frequency and inductor's inductance?

 

[audioguru2]

Since the 4047 is in the 4000 series, I assume it doesn't work down to 1.5v? I thought their voltage range was just 3v-15v?
Also, I thought that with boost converters, if the duty cycle is higher than 50%, the efficiency can be increased by matching the duty cycle to the frequency and inductor's inductance?

Only 1.5V? I missed that.
A CD4047 works down to 3V and a 74HC4047 works down to 2V.
My solar garden lights work down to 1V with only two transistors. Their output current is low.
The amount of switching time and the amount of supply current that is wasted determines the efficiency.

 

[Hero999]

I have an idea but it will fail your 'not more than 1A' requirement as it doesn't have current limiting.

Are you aware that 5V@1A out means [email protected] in?

You've really set some impossible requirements, just buy an IC and be done with it.

Here's my idea but it won't be very efficient, LTSpice says the output is 4.5V@200mA with 1.5V in.

View attachment 40832

 

[gogo2520]

hello
That is a heavy Schottky Barrier diode, it can handle 40 volts 7 amps, you think it will need something that heavy? Other wise besides finding the coil it should be easy enough.
                                                              have fun
                                                                gogo

heres the data sheet for the mbr745

ds23007.pdf

 

[Hero999]

It's just a simulation.

The peak current will be at least 6.67 times the output  current

Although components can pass higher peak currents than the continuous rating the voltage drop will be hgiher so it's a good idea to overrate.

Use a transistor with a higher current rating than the one used for Q2 in the simulation , for 200mA out, use one with a 1.5A rating, for 0.5A out use a transistor rated to at least 3.5A and the same goes for the diode. You'll also probably need a higher gain transistor than the one in the simulation.

I'd recommend the ZTX689B because it has a very high gain, a low saturation voltage and can handle high peak currents 8A.
http://www.rapidonline.com/netalogue/specs/81-0228.pdf

EDIT:
I forgot to add a link to the circuit my design is based on: the Joule Thief.

http://www.joulethief.com/kit.php

Do you know that  a Joule Thief can also be used as a bootstrap to enable DC-DC converter ICs which require over 1.5V to start at lower voltages?

Here's an example showing how to use the other Joule Thief circuit as a blocking oscillator:

http://www.electro-tech-online.com/electronic-projects-design-ideas-reviews/92584-using-joule-thief-bootstap.html

Note that I no longer post on the above forum so don't reply to me there or send me a PM; I won't respond.

 

[pyrohaz1]

Hi there hero! thank you very much for your help and reply. If you re-read my spec, it was 0.2A our, not 1A at 5V. I just wondered if the input current could be kept below 1A because thats the maximum current supplyable by a D cell alkaline battery!

Now, I definately do like the switching circuit that you designed for me, what is the switching frequency? (I didn't put this into the specification but i did want it to hopefully be above >25kHz, out of the audible range BUT thats not important .

Thank you very much for the help, would you possibly be able to run through with me how you designed that circuit please for future references?

Thanks again!

 

[Hero999]

A 1.5V D cell can provide a lot more than an Amp which is good because it isn't possible to convert 1.5V to 5V 0.5A without drawing more than 1A.

I can't remember what the frequency was when I simulated it. I remember adjusting the component values because it was very low with the original Joule Thief circuit.

 

[audioguru2]

A alkaline D cell is 1.5V only when it is brand new. When it is supplying 1A or more then its voltage keeps dropping.

It can supply 1A for 4 hours when its voltage will have dropped to 1.0V.
It can supply 5A for about 20 minutes when its voltage will have dropped to 1.0V.

 

[pyrohaz1]

Wow! 4 Hours! that would be brilliant, I only expected 2!

So as Hero said, how about:

Joule Thief bootstrapping a 555 Timer, acting as a PWM oscillator at a given frequency, with a switching output?

That circuit would be ok right because I know the joule thief can supply atleast 20mA (to light an LED)

 

[Hero999]

Yes, the capacity is inversely proportional to the current drawn, the higher the current the lower the capacity.
http://data.energizer.com/PDFs/E95.pdf

The oscillator part of the circuit consists of two inverting amplifiers formed by Q1 and Q2 with positive feedback. The inductor and capacitor prodce phase shift and oscillation. The high voltage pulses are rectified and smoothed by a capacitor. When the zener's breakdown is exceeded Q1 is turned permanently on causing the oscillator to stop; this regulates the voltage at around 5V.

I think you should ditch the D cell idea.

You'd probably be better off using three AA cells and a boost converter. It's more efficient to step-up 4.5V to 5 V than it is to boost 1.5V to 5V.

Three AA cells will last for four hours down to 0.8V per cell (2.4V total) with a current draw of 400mA. Of course the current draw will start low and increase as the battery discharges so it's difficult to predict how log it will last for. Another option is to use six AAA cells and a low drop-out buck converter.

I said nothing about using a 555 timer, that was your idea. I think you should use a real SMPs IC, it's not like it's expensive and might save you money in destroyed components. ;D

http://data.energizer.com/PDFs/e91.pdf